MATH 308 Lecture 25

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Lecture Notes

Convolution Function

Defined

$(f*g)(t)=\underset{0}{\overset{t}{\int }}f(t-v)g(v)\mathrm{d}v=\underset{0}{\overset{t}{\int }}f(u)g(t-u)\mathrm{d}u$

• It is associative, so $f*g=g*f$
• It is distributive, so $(f*(g(t)+ah(t)))(t)=(f*g)(t)+a(f*h)(t)$
• $f*0(t)=0$

Example

$f(t)={\mathrm{e}}^t$, and $g(t)=\sin (3t)$:

$(f*g)(t)=\underset{0}{\overset{t}{\int }}{\mathrm{e}}^{t-v}\sin 3v\mathrm{d}v=\underset{0}{\overset{t}{\int }}{\mathrm{e}}^t\sin (3(t-v))\mathrm{d}v$

$f(t)=t^2$, and $g(t)=1$

$(f*g)(t)=\underset{0}{\overset{t}{\int }}{v}^2\cdot 1\mathrm{d}v=\frac{t^3}{3}$

Theorem 6.6.1

${ℒ}\{f*g\}={ℒ}\left\{f\right\}{ℒ}\left\{g\right\}$

So finding the inverse Laplace transform of $\frac{1}{(s-7{)}^9(s^2+4)}$ is equivalent to finding the convolution $\frac{(t^8{\mathrm{e}}^{7t})*(\sin 2t)}{8!\cdot 2}$

Exercises

1. ${{ℒ}}^{-1}\left\{\frac{1}{(s-1)(s^2+9)}\right\}={\mathrm{e}}^t*\frac{\sin 3t}{3}$
2. ${ℒ}\{g(t)\}=G(s)$, $F(s)=G(s)\frac{1}{s^2+1}={ℒ}\{g(t)*\sin t\}$

Exercise 3

Find Laplace transforms of

1. $f(t)=\underset{0}{\overset{t}{\int }}(t-v{)}^3\sin 2v\mathrm{d}v=t^3*\sin 2t$, so ${ℒ}\{f(t)\}=\frac{6}{s^4}\frac{2}{s^2+4}=\frac{12}{s^4(s^2+4)}$
2. $g(t)=\underset{0}{\overset{t}{\int }}{\mathrm{e}}^{v-t}\cos v\mathrm{d}v={\mathrm{e}}^{-t}*\cos t$, so ${ℒ}\{g(t)\}=\frac{s}{s^2+1}\frac{1}{s+1}=\frac{s}{(s^2+1)(s+1)}$

Solving Differential Equations

Find the solution to $y^{″}+{\omega }^{2}y=g(t)$ when $y(0)=0$ and $y^{\prime }(0)=1$.

Take Laplace transform of both sides:

$s^2{ℒ}\left\{y\right\}-y^{\prime }(0)-sy(0)+{\omega }^2{ℒ}\left\{y\right\}={ℒ}\{g(t)\}$

Solve for ${ℒ}\left\{y\right\}$:

${ℒ}\left\{y\right\}=\frac{{ℒ}\{g(t)\}}{s^2+{\omega }^2}+\frac{1}{s^2+{\omega }^2}$

So the solution is

$y=g(t)*\frac{\sin \omega t}{\omega }+\frac{\sin \omega t}{\omega }$

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Find the solution to $y^{″}+3y^{\prime }+2y=\cos at$ when $y(0)=1$ and $y^{\prime }(0)=0$

We can solve this using two methods: undetermined coefficients and variation of parameters. Let's try to use Laplace transforms:

$s^2{ℒ}\left\{y\right\}-y^{\prime }(0)-sy(0)+3{ℒ}\left\{y\right\}-3y(0)+2{ℒ}\left\{y\right\}=\frac{s}{s^2+a^2}$

Solve for ${ℒ}\left\{y\right\}$:

$\begin{array}{rl}{ℒ}\left\{y\right\}& =\frac{s}{(s^2+a^2)(s^2+3s+2)}+\frac{s+3}{s^2+3s+2}\\ & =\left(\frac{s}{s^2+a}\right)\left(\frac{1}{s^2+3s+2}\right)+\frac{s+3}{(s+2)(s+1)}\\ & ={ℒ}\{\cos at\}{ℒ}\left\{{\mathrm{e}}^{-2t}\right\}{ℒ}\left\{{\mathrm{e}}^{-t}\right\}+\frac{s+3}{(s+2)(s+1)}\\ & ={ℒ}\{\cos at\}{ℒ}\left\{{\mathrm{e}}^{-2t}\right\}{ℒ}\left\{{\mathrm{e}}^{-t}\right\}-\frac{1}{s+2}+2\frac{1}{s+1}\end{array}$

And then take the inverse Laplace transform:

(out of time)